198. House Robber

题意：You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

思路：简单的DP问题，找到对应的递推式就好了，我开始的想法是r[i] = max(r[i-2],r[i-3])+nums[i]，这样的问题是其实是把递推式复杂化了一点，需要更远的值来递推求解，就需要更多的数据空间，所以我声明了一个数组，还在前面加了三个0来方便初始化，具体见代码：

class Solution {public:    int rob(vector<int>& nums) {        int m = nums.size();        vector<int> result (m+3, 0);        for(int i=3;i<m+3;i++){            result[i] = max(result[i-2],result[i-3])+nums[i-3];        }        return max(result[m+1],result[m+2]);    }};

其实最简单的递推式应该是：
f(0)=nums[0]
f(1)=max(num[0],num[1])
f(k)=max(f(k−2)+nums[k],f(k−1))
这样就只需要常数额外空间就可以求解了：

class Solution:    def rob(self, nums):        last, now = 0, 0        for i in nums: last, now = now, max(last + i, now)        return now